# Pinter 10.E.5: Let $a$ and $b$ commute. Prove: There is an element $c$ in $G$ whose order is $lcm(m,n)$.

This is 10.E.5 from Pinter.

Let $$a$$ and $$b$$ be elements of a group $$G$$.

Let $$\text{ord}(a) = m$$ and $$\text{ord}(b) = n$$.

Let $$a$$ and $$b$$ commute.

Prove:

There is an element $$c$$ in $$G$$ whose order is $$lcm(m,n)$$.

(HINT: Use 10.E.4 together with 10.D.3. Let $$c = a^i b$$ where $$a^i$$ is a certain power of $$a$$.

Here is 10.E.4:

Let $$a$$ and $$b$$ commute.

If $$m$$ and $$n$$ are relatively prime, then $$ord(ab) = mn$$.

(HINT: Use 10.E.2)

Here is 10.D.3:

If $$ord(a) = km$$, then $$ord(a^k) = m$$.

Here's the start of one approach.

As Pinter suggests, let

$$c = a^i b$$

Let's start from what is to be proven:

$$ord(c) = lcm(m,n)$$

$$ord(a^i b) = lcm(m,n)$$

$$(a^i b)^{lcm(m,n)} = e$$

$$(a^i)^{lcm(m,n)} b^{lcm(m,n)} = e$$

$$a^{i lcm(m,n)} b^{lcm(m,n)} = e$$

$$a^{i lcm(m,n)} = b^{-lcm(m,n)}$$

$$a^i a^{lcm(m,n)} = b^{-lcm(m,n)}$$

$$a^i = b^{-lcm(m,n)} a^{-lcm(m,n)}$$

$$a^i = (ba)^{-lcm(m,n)}$$

In other words, we're aiming to solve for $$a^i$$.

However, this doesn't seem too helpful as it simplifies to $$e$$.

Moreover, we haven't used 10.E.5 or 10.D.3.

It seems that this part of the hint:

$$c = a^i b$$ where $$a^i$$ is a certain power of $$a$$

is suggesting that we find an $$a^i$$ such that $$gcd(a^i,b) = 1$$.

And once that's done, apply 10.E.4:

$$ord(a^i b) = ord(a^i) n$$

And hopefully, together with 10.D.3, change that to:

$$ord(a^i b) = lcm(m,n)$$

This question has been discussed in a few places on math.se.

Here's one approach.

The first part of the answer appears to prove 10.E.4 and solves the case for when $$m$$ and $$n$$ are relatively prime.

The second part of the answer does not appear to apply 10.D.3 as recommended by Pinter.

There are many approaches illustrated on this post.

However, none appear to use the path suggested by Pinter.

The answer here essentially restates part of Pinter's hint.

Is there a clear approach which uses Pinter's suggestions to use 10.D.3, 10.E.4, and only what has been covered up to that point in the book?

# Bill's answer converted to Pinter's notation

• This is a conversion of the proof in Bill's answer to the notation in Pinter.
• It also makes many of the implied steps explicit.
• This will hopefully make the proof more familiar and digestible to folks coming to Pinter's book and abstract algebra perhaps for the first time.

We induct on

$$ord(a) ord(b)$$

## Part 1

If

$$ord(a) ord(b) = 1$$

the only possible values are:

$$ord(a) = 1$$

$$ord(b) = 1$$

so

$$lcm(ord(a),ord(b)) = 1$$

$$ord(c) = 1$$

which means

$$c^1 = e$$

thus

$$c = e$$

## Part 2

Now let's consider the case where

$$ord(a)ord(b) != 1$$

Split off $$p^k$$ in $$ord(a)$$, $$ord(b)$$. I.e.

$$ord(a) = xP$$ $$ord(b) = yP'$$

where

$$P'|P$$

$$P = p^k > 1 \tag{7}$$

$$p$$ is prime and $$p∤x,y$$.

Then, by 10.D.3:

$$ord(a^P) = x \tag{1}$$ $$ord(b^{P'}) = y \tag{2}$$

$$ord(a^x) = P \tag{3}$$ $$ord(b^y) = P' \tag{4}$$

By induction, there is a $$c$$ with

$$ord(c) = lcm(x,y)$$

Substitute (1) and (2):

$$ord(c) = lcm(ord(a^P),ord(b^{P'})) \tag{5}$$

$$c = a^P b^{P'} \tag{6}$$

Substitute (6):

$$ord(a^P b^{P'}) = lcm(ord(a^P),ord(b^{P'}))$$

So:

$$ord(a^x c)$$

$$lcm(ord(a^x),ord(c))$$

Substitute (3) and (5):

$$lcm(P, lcm(ord(a^P), ord(a^{P'})))$$

$$lcm(P, lcm(x,y))$$

Since $$P = p^k$$ where $$p$$ is prime and $$p$$ is coprime to $$x$$ and $$y$$:

$$P lcm(x,y)$$

$$lcm(xP, yP')$$

$$lcm(ord(a), ord(b))$$

Thus:

$$ord(a^x c) = lcm(ord(a), ord(b))$$

# Pinter's approach

This is a start of an approach that closely follows Pinter's hint.

(I have this written up in a text file and didn't have time to format as LaTeX.)

Let ord(a) = m.
Let ord(b) = n.

Let i = gcd(m,n).           (1)

Let ord(a) = k i            (2)
========================================
Consider: ord(a^i b)

If ord(a^i) and ord(b) are relatively prime then by 10.E.4:

ord(a^i b) = ord(a^i) ord(b)        (3)
========================================
By eq (2) and 10.D.3:

ord(a^i) = k                (4)

Eq (2):

ord(a) = k i

Substitute (4):

ord(a) = ord(a^i) i

Solve for ord(a^i):

ord(a^i) = ord(a)/i         (5)
========================================
Eq (3):

ord(a^i b) = ord(a^i) ord(b)

Substitute (5):

ord(a^i b) = ord(a)/i ord(b)

ord(a^i b) = ord(a) ord(b) / i

Substitute (1):

ord(a^i b) = ord(a) ord(b) / gcd(m,n)

ord(a^i b) = ord(a) ord(b) / gcd(ord(a),ord(b))

ord(a^i b) = lcm(a,b)
----------------------------------------


The only remaining part left to do is to handle the case where ord(a^i) and ord(b) are not relatively prime.

• See also this post. – Dietrich Burde Jul 28 '19 at 10:43
• Yes, that's the approach used in the proof I gave here (E.4 is used in the final line) – Bill Dubuque Jul 28 '19 at 15:29
• No they're equivalent - just apply it to the abelian sugroup generated by $a,b$ (or note that my proof only needs that $a,b$ commute). – Bill Dubuque Jul 28 '19 at 19:57
• Yes, it is used in a few places without explicit mention. I just added a link to a proof of E.4 there. – Bill Dubuque Jul 28 '19 at 20:35
• I added a link for the proof of D.3 too. – Bill Dubuque Jul 28 '19 at 20:53

This is easily done by exploiting prime factorisations. With your notations, write

$$m= p_1^{k_1}\dots p_s^{k_s}; n = p_1^{l_1}\dots p_s^{l_s}$$

We can express $$lcm(m,n)$$ as

$$p_1^{\min\{k_1, l_1\}}\dots p_s^{\min\{k_s,l_s\}}$$

If we can find elements $$c_i$$ with $$ord(c_i) = p_i^{\min\{k_i,l_i\}}$$, then the element you are looking for is $$c=\prod_i c_i$$ because the order is multiplicative on commutating elements with coprime orders.

Put $$c_i =\begin{cases} a^{m/p^{k_i}} \quad k_i\leq l_i \\ b^{n/p^{l_i}} \quad l_i < k_i \end{cases}$$

and you see we are done.

• That's essentially the same as the proof I linked in the comments, except I do the induction explicitly (vs. implicitly above). – Bill Dubuque Jul 30 '19 at 13:10